[Math] Convolution of tempered distributions where one has compact support.

Question

For $u\in\mathcal E'(\mathbb R^n)$ and $v\in\mathcal S'(\mathbb R^n)$, we defined $u\ast v$ by $\langle u\ast v, \phi\rangle = \langle v, \check u \ast \phi \rangle$ for all $\phi\in\mathcal S(\mathbb R^n)$ (we showed before that this definition makes sense).

I shall now show, that $\langle u\ast v, \phi\rangle = \langle u, \phi\ast \check v \rangle$ for all $\phi\in\mathcal S(\mathbb R^n)$ holds aswell

I reduced that to the case where $v$ is a smooth function which additionally is of polynomial growth but I don't know how to proceed, any hints?

Answer 1

Notation: For a fixed function $f$ defined on $\mathbb{R}^{n}$ and $y\in\mathbb{R}^{n}$, we write $\tilde{f}(x)=f(-x)$ and $(\tau^{y}f)(x)=f(x-y)$.

I am assuming that you are familiar with the following result.

Theorem. For $\phi\in\mathcal{S}(\mathbb{R}^{n})$, the tempered distribution $u\ast\phi$ coincides with a $C^{\infty}(\mathbb{R}^{n})$ function $\langle{u,\tau^{x}\tilde{\phi}}\rangle$ satisfying the growth estimates $$\left|\partial^{\alpha}\langle{u,\tau^{x}\tilde{\phi}}\rangle\right|\leq C_{\alpha}(1+\left|x\right|)^{N_{\alpha}},\qquad\forall x\in\mathbb{R}^{n}$$ for all multi-indices $\alpha$. Moreover, if $u$ has compact support, then $u\ast\phi$ coincides with a Schwartz function.

And I leave it to you to prove the following lemma.

Lemma. Let $\phi\in\mathcal{S}(\mathbb{R}^{n})$ be such that $\int\phi=1$, and set $\phi_{\epsilon}:=\epsilon^{-n}\phi(\epsilon^{-1}\cdot)$. Then $$f\ast\phi_{\epsilon}\rightarrow f \text{ in }\mathcal{S}(\mathbb{R}^{n}), \qquad\forall f\in\mathcal{S}(\mathbb{R}^{n})$$ as $\epsilon\downarrow 0$.

By duality, we see that for any tempered distribution $u$, $$\langle{u\ast\phi_{\epsilon},f}\rangle=\langle{u,f\ast\widetilde{\phi_{\epsilon}}}\rangle=\langle{u,f\ast(\tilde{\phi})_{\epsilon}}\rangle\rightarrow \langle{u,f}\rangle$$ as $\epsilon\downarrow 0$. Whence, $u\ast\phi_{\epsilon}\rightarrow u$ in $\mathcal{S}'(\mathbb{R}^{n})$ as $\epsilon\downarrow 0$.

Define linear functionals $w_{1}$ and $w_{2}$ on $\mathcal{S}(\mathbb{R}^{n})$ respectively by $$\langle{w_{1},\psi}\rangle:=\langle{u,\tilde{v}\ast\psi}\rangle, \quad \langle{w_{2},\psi}\rangle:=\langle{v,\tilde{u}\ast\psi}\rangle \qquad\forall \psi\in\mathcal{S}(\mathbb{R}^{n})$$ If I read your post correctly, you know how to verify that $w_{1}$ and $w_{2}$ are tempered distributions, so I will omit that detail. If we can show that $w_{1}\ast\phi=w_{2}\ast\phi$ in $\mathcal{S}'(\mathbb{R}^{n})$, then by replacing $\phi$ with $\phi_{\epsilon}$ and letting $\epsilon\downarrow 0$, we conclude that $w_{1}=w_{2}$ in the sense of tempered distributions.

Fix $\phi\in\mathcal{S}(\mathbb{R}^{n})$ and observe that for any $\psi\in\mathcal{S}(\mathbb{R}^{n})$, \begin{align*} \langle{w_{1}\ast\phi,\psi}\rangle=\langle{w_{1},\tilde{\phi}\ast\psi}\rangle&=\langle{u,\tilde{v}\ast(\tilde{\phi}\ast\psi)}\rangle\\ &=\langle{u,(\tilde{v}\ast\psi)\ast\tilde{\phi}}\rangle\\ &=\int_{\mathbb{R}^{n}}(u\ast\phi)(x)(\tilde{v}\ast\psi)(x)\mathrm{d}x\\ \end{align*} where we use the commutativity of the convolution of ordinary functions, and the associativity of the convolution of a function and a distribution.

Similarly, \begin{align*} \langle{w_{2}\ast\phi,\psi}\rangle=\langle{v,\tilde{u}\ast(\tilde{\phi}\ast\psi)}\rangle=\langle{v,(\tilde{u}\ast\tilde{\phi})\ast\psi}\rangle&=\langle{v\ast\tilde{\psi},\tilde{u}\ast\tilde{\phi}}\rangle\\ \end{align*} But \begin{align*} \int_{\mathbb{R}^{n}}(u\ast\phi)(x)(\tilde{v}\ast\psi)(x)\mathrm{d}x=\int_{\mathbb{R}^{n}}\langle{u,\tau^{x}\tilde{\phi}}\rangle\langle{v,\widetilde{\tau^{x}\tilde{\psi}}}\rangle\mathrm{d}x&=\int_{\mathbb{R}^{n}}\langle{u,\tau^{x}\tilde{\phi}}\rangle\langle{v,\tau^{-x}\psi}\rangle\mathrm{d}x\\ &=\int_{\mathbb{R}^{n}}\langle{u,\tau^{-x}\tilde{\phi}}\rangle\langle{v,\tau^{x}\psi}\rangle\mathrm{d}x\\ &=\int_{\mathbb{R}^{n}}\langle{u,\widetilde{\tau^{x}\phi}}\rangle\langle{v,\tau^{x}\psi}\rangle\mathrm{d}x\\ &=\langle{v\ast\tilde{\phi},\tilde{u}\ast\tilde{\psi}}\rangle \end{align*} This completes the proof.